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3 years ago

Consider the nuclear equation below.

Chemistry

2 answers:

Dovator [93]3 years ago

8 0

Answer:

its NOT C

Explanation:

Kamila [148]3 years ago

3 0

Answer: 228/88 Ra

Explanation:

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3 years ago

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An inventor claims to have devised a closed cyclic engine which exchanges heat with cold and hot reservoirs at 25 and 350 ◦C, re

kati45 [8]

Explanation:

The given data is as follows.

$T_{c} = 25^{o}C$ , $T_{h} = 350^{o}C$

Work produced per kJ of heat extracted from hot reservoir = 0.45 kJ = Efficiency

If the device is Carnot cycle then its efficiency will be maximum and its value will be equal to $[1 - (\frac{T_{c}}{T_{h}} )]$

Using this relation we will calculate the efficiency as follows.

Efficiency = $[1 - (\frac{T_{c}}{T_{h}} )]$

= $1 - (\frac{25}{350})$

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Hence, it means that this type of device is possible and the claim is also believable.

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2 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

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3 years ago

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